What is the equation of the line tangent to #f(x)=6x^2 + 4x - 9 # at #x=-1#?

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Answer 1 · 2016-10-09T21:06:30

#y = -8x - 12#

Begin by computing the first derivative, #f'(x)#:

#f'(x) = 12x + 4#

The slope, m, of the tangent line is, #f'(-1)#:

#m = f'(-1)#
#m = 12(-1) + 4#
#m = -8#

Obtain the y coordinate by evaluating #f(-1)#:

#y = f(-1)#
#y = 6(-1)² + 4(-1) - 9#
#y = -7#

Using the point-slope form of the equation of a line:

#y - y_1 = m(x - x_1)#

Substitute -7 for #y_1#, -1 for #x_1# and -8 for #m#

#y - -4 = -8(x - -1)#

#y + 4 = -8x - 8#

#y = -8x - 12#